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| 1. |
If no root of the equation x2-px+1=0is real ,then find the value of p |
| Answer» Replacing p with a,Given, {tex}x^2 - ax +1=0{/tex}Comparing given equation with\xa0{tex}Ax^2 +Bx +C{/tex}, we get\xa0{tex}A = 1,\\ B = -a\\ and\\ C = 1{/tex}The given equation will has no real roots, D <\xa00{tex}\\Rightarrow{/tex}\xa0{tex}B^2\xa0- 4AC <\xa00{/tex}{tex}\\Rightarrow{/tex}\xa0(-a)2\xa0- 4(1)(1) < 0{tex}\\Rightarrow{/tex}\xa0{tex}a^2\xa0-4 < 0{/tex}{tex}\\Rightarrow{/tex}\xa0a2\xa0< 4\xa0{tex}\\Rightarrow{/tex}\xa0a <\xa0{tex}\\sqrt{4}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0-2 < a < 2\xa0 | |